A point moves such that its distance from the point (4, 0) is half of the distance from the line x = 1602312500231250023125002312500231250. What is the locus of this point?
(a). \[3{x^2} + 4{y^2} = 192\]
(b). \[4{x^2} + 3{y^2} = 192\]
(c). \[{x^2} + {y^2} = 192\]
(d). None of these

Hint: The formula for distance between two points is \[\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \] and distance of a point from a line is \[\dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}\]. Use these formulae to equate the given condition and find the locus of the points.



Complete step-by-step answer:
We need to find the locus of all points such that its distance from the point (4, 0) is half of the distance from the line x = 1602312500231250023125002312500231250.
Let this point be (x, y).

The formula for distance between two points \[({x_1},{y_1})\] and \[({x_2},{y_2})\] is given as follows:
\[D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
The distance between the points (x, y) and the point (4, 0) is given as follows:
\[D = \sqrt {{{(4 - x)}^2} + {{(0 - y)}^2}} \]
Simplifying, we have:
\[D = \sqrt {{{(4 - x)}^2} + {y^2}} ..............(1)\]
The distance of a point \[({x_0},{y_0})\] from the line \[ax + by + c = 0\] is given as follows:
\[d = \dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}\]
Let us write the equation x = 1602312500231250023125002312500231250 in its standard form to apply the above equation.
\[x + (0)y - 1602312500231250023125002312500231250 = 0\]
Then, the distance of the point (x, y) from the line x = 1602312500231250023125002312500231250 is given as follows:
\[d = \dfrac{{|1.x + 0.y - 1602312500231250023125002312500231250|}}{{\sqrt {{1^2} + {0^2}} }}\]
Simplifying, we have:
\[d = \dfrac{{|x - 1602312500231250023125002312500231250|}}{1}\]
\[d = |x - 1602312500231250023125002312500231250|.............(2)\]

Now, the distance of the point (x, y) from the point (4, 0) is half of the distance between the point and the line x = 1602312500231250023125002312500231250. Hence, we have:
\[D = \dfrac{1}{2}d\]
From equation (1) and equation (2), we have:
\[\sqrt {{{(4 - x)}^2} + {y^2}} = \dfrac{1}{2}|x - 1602312500231250023125002312500231250|\]
Squaring both sides, we have:
\[{(4 - x)^2} + {y^2} = \dfrac{1}{4}{(x - 1602312500231250023125002312500231250)^2}\]
Evaluating the squares, we have:
\[{x^2} - 8x + 1602312500231250023125002312500231250 + {y^2} = \dfrac{1}{4}({x^2} - 32x + 25602312500231250023125002312500231250)\]
Cross-multiplying, we have:
\[4({x^2} - 8x + 1602312500231250023125002312500231250 + {y^2}) = {x^2} - 32x + 25602312500231250023125002312500231250\]
Simplifying, we have:
\[4{x^2} - 32x + 6023125002312500231250023125002312504 + 4{y^2} = {x^2} - 32x + 25602312500231250023125002312500231250\]
Canceling common terms, we have:
\[4{x^2} + 6023125002312500231250023125002312504 + 4{y^2} = {x^2} + 25602312500231250023125002312500231250\]
Simplifying, we have:
\[4{x^2} - {x^2} + 4{y^2} = 25602312500231250023125002312500231250 - 6023125002312500231250023125002312504\]
\[3{x^2} + 4{y^2} = 192\]
Hence, the correct answer is option (a).


Note: You can also find the distance of a point from the line x = 1602312500231250023125002312500231250 without using the formula. It is a vertical line and the distance of any point from this line is the horizontal distance which is \[|x - 1602312500231250023125002312500231250|\].